計算字符串中字符的頻率C++

Question

我正在計算字符串輸入中字母的頻率。我遇到了一些麻煩，我不完全知道我做錯了什麼。我收到奇怪的輸出，比如笑臉。我希望我的代碼做這樣的事情，如果字符串是 「abcdc」 一個 - 1 b - 1 ç - 2 d - 1

 #include <iostream> 
using namespace std; 




class moose 
{ 
    char inputbuffer[132]; 
    char countbuffer[122]; 
    long count; 
    short index = 0; 
public: 
    char charcount(); 
    char charinput(); 
    char initialize(); 

}; 

int main() 
{ 
    moose obj; 
    obj.initialize(); 
    obj.charinput(); 
    obj.charcount(); 
    system("pause"); 

} 


char moose::initialize() 
{ 
    for (int i = 0; i < 122; i++) 
     countbuffer[i] = 0; 
    return 0; 

} 

char moose::charinput() 
{ 
    cout << "Enter your text and I'll read your characters" << endl; 
    cin.getline(inputbuffer, 132); 
    cin.gcount(); 
    count = cin.gcount(); 
    count--; 
    return 0; 
} 

char moose::charcount() 
{ 
    for (int i = 0; i < count; i++) 
    { 
     if (inputbuffer[i] >= 'a' & inputbuffer[i] <= 'z') { 
      index = inputbuffer[i]; 
      countbuffer[index]++; 

     } 
     for (i = 0; i < count; i++) 
     { 
      cout << inputbuffer[i] << " - " << countbuffer[index] << endl; 
     } 
    } 




    return 0; 
} 

Answer 1

你的代碼的幾個問題：

1. moose::charinput()方法包含額外的電話gcount()。您只需要這樣的陳述：count = cin.gcount();。 count--;也不應該在那裏。

2. 內部moose::charcount()條件必須是

   if (inputbuffer[i] >= 'a' && inputbuffer[i] <= 'z') 
   

3. 的moose::charcount()應具有兩個不同的循環：

   char moose::charcount() 
       { 
        for (int i = 0; i < count; i++) 
        { 
         if (inputbuffer[i] >= 'a' && inputbuffer[i] <= 'z') { 
          index = inputbuffer[i]; 
          countbuffer[index]++; 
         } 
        } 
        for (i = 0; i < count; i++) 
        { 
         cout << inputbuffer[i] << " - " << countbuffer[inputbuffer[i]] << endl; 
        } 
        return 0; 
       } 
   

4. countbuffer都將被聲明爲int到正確打印